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How do you find the maximum mass of a limiting reactant?

Published in Stoichiometry 3 mins read

The maximum mass of the product formed when a limiting reactant is completely consumed can be determined using stoichiometry. You don't directly find the "maximum mass" of the limiting reactant; it's all consumed. Instead, you calculate the maximum mass of product it can produce. Here's how:

  1. Balance the Chemical Equation: Ensure the chemical equation is balanced. This provides the crucial mole ratios needed for stoichiometric calculations. A balanced equation has the same number of each type of atom on both sides of the equation. For example:

    2H₂ + O₂ → 2H₂O

  2. Determine the Moles of Each Reactant: Convert the given mass of each reactant to moles using its molar mass. The molar mass is found on the periodic table.

    • Moles = Mass (g) / Molar Mass (g/mol)

  3. Identify the Limiting Reactant: The limiting reactant is the reactant that produces the least amount of product.

    • Choose one of the products.
    • For each reactant, use the mole ratio from the balanced equation to calculate the moles of the chosen product that could be formed if that reactant was completely consumed.
    • The reactant that yields the least amount of product is the limiting reactant.

  4. Calculate the Theoretical Yield (Maximum Mass of Product): Using the moles of product calculated from the limiting reactant in the previous step, convert this to grams using the product's molar mass. This is the maximum mass of product that can be formed.

    • Mass (g) = Moles × Molar Mass (g/mol)

Example:

Let's say we have the reaction:

N₂ + 3H₂ → 2NH₃

And we start with 28g of N₂ and 9g of H₂. What is the maximum mass of NH₃ that can be formed?

  1. Balanced Equation: Already balanced above.

  2. Moles of Reactants:

    • N₂: 28 g / 28 g/mol = 1 mole
    • H₂: 9 g / 2 g/mol = 4.5 moles

  3. Limiting Reactant: Let's choose NH₃ as the product to compare.

    • From N₂: 1 mole N₂ * (2 moles NH₃ / 1 mole N₂) = 2 moles NH₃
    • From H₂: 4.5 moles H₂ * (2 moles NH₃ / 3 moles H₂) = 3 moles NH₃

    Since N₂ produces less NH₃ (2 moles) than H₂ (3 moles), N₂ is the limiting reactant.

  4. Theoretical Yield (Maximum Mass of NH₃):

    • Moles of NH₃ formed = 2 moles (from the limiting reactant, N₂)
    • Mass of NH₃ = 2 moles * 17 g/mol = 34 g

Therefore, the maximum mass of NH₃ that can be formed is 34g.

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